Important Derivations

Steps

1. 1

   Consider a dipole with charges +q and −q separated by distance 2l

2. 2

   Take a point P on the axial line at distance r from centre

3. 3

   E due to +q: E₁ = kq/(r−l)² (towards P)

4. 4

   E due to −q: E₂ = kq/(r+l)² (away from P)

5. 5

   E_net = E₁ − E₂ = kq[1/(r−l)² − 1/(r+l)²]

6. 6

   Simplify and apply r >> l: E_axial = 2kp/r³

Steps

1. 1

   Take point P on equatorial line at distance r from centre

2. 2

   E₁ = E₂ = kq/(r² + l²) — magnitudes are equal

3. 3

   Vertical components cancel; horizontal components add up

4. 4

   E_net = 2E₁ cosθ where cosθ = l/√(r²+l²)

5. 5

   Substitute and simplify: E_eq = kp/(r²+l²)^(3/2)

Steps

1. 1

   Two large parallel plates of area A, separated by distance d, with charges +Q and −Q

2. 2

   Surface charge density σ = Q/A

3. 3

   Electric field between plates: E = σ/ε₀ = Q/(ε₀A) (using Gauss's law)

4. 4

   Potential difference: V = E × d = Qd/(ε₀A)

5. 5

   Capacitance C = Q/V = ε₀A/d

Steps

1. 1

   When charge q has already accumulated on capacitor, potential V = q/C

2. 2

   Small work done to transfer additional charge dq against this potential: dW = V dq = (q/C) dq

3. 3

   Total work done to charge capacitor from 0 to Q: U = ∫₀Q (q/C) dq = [q²/2C]₀Q = Q²/2C

4. 4

   Since Q = CV: U = (CV)²/2C = ½CV²

5. 5

   Also U = ½QV — all three forms are equivalent

6. 6

   This energy is stored in the electric field between the plates

Steps

1. 1

   Each element dl of the circular loop is perpendicular to r

2. 2

   dB = (μ₀/4π) × Idl/R² (since sinθ = 1 for every element)

3. 3

   All dB elements point in same direction (into or out of loop plane)

4. 4

   Integrate: B = (μ₀/4π) × I/R² × ∮dl = (μ₀/4π) × I × 2πR/R²

5. 5

   B = μ₀I/2R

Steps

1. 1

   Consider a rectangular Amperian loop ABCDA of length l inside solenoid

2. 2

   Only side AB (inside solenoid) contributes to ∮B·dl

3. 3

   ∮B·dl = Bl (contributions from BC, CD, DA are zero)

4. 4

   Total enclosed current = nIl (n turns per unit length, l = length)

5. 5

   Applying Ampere's Law: Bl = μ₀nIl → B = μ₀nI

Steps

1. 1

   A rod of length l moves with velocity v perpendicular to magnetic field B

2. 2

   Free electrons experience force: F = qv × B (upward for + charges)

3. 3

   Electrons accumulate at one end, creating potential difference

4. 4

   Equilibrium: electric force qE = magnetic force qvB → E = vB

5. 5

   EMF = E × l = Blv

Steps

1. 1

   Work done against back-EMF when current increases by dI: dW = εI dt

2. 2

   Back-EMF: ε = L(dI/dt)

3. 3

   dW = L(dI/dt) × I × dt = LI dI

4. 4

   Total work (energy stored): U = ∫₀ᴵ LI dI = ½LI²

Steps

1. 1

   Consider object AB beyond C (centre of curvature) of concave mirror

2. 2

   Two rays from tip B: parallel to principal axis (reflects through F), and through C

3. 3

   Image A'B' formed between F and C

4. 4

   Using similar triangles: A'B'/AB = A'P/AP ...(i)

5. 5

   Also: A'B'/AB = A'F/FP ...(ii)

6. 6

   From (i) and (ii): A'F/FP = A'P/AP

7. 7

   Substituting with sign convention and simplifying: 1/v + 1/u = 1/f

Steps

1. 1

   Apply refraction formula at first surface (radius R₁): n₂/v₁ − n₁/u = (n₂−n₁)/R₁

2. 2

   Image I₁ acts as virtual object for second surface

3. 3

   Apply refraction at second surface (radius R₂): n₁/v − n₂/v₁ = (n₁−n₂)/R₂

4. 4

   Add both equations (v₁ cancels): n₁/v − n₁/u = (n₂−n₁)[1/R₁ − 1/R₂]

5. 5

   Divide by n₁: 1/v − 1/u = (n−1)[1/R₁ − 1/R₂] = 1/f

Steps

1. 1

   Centripetal force = Coulomb force: mv²/r = Ze²/4πε₀r²

2. 2

   Bohr's quantisation condition: mvr = nh/2π

3. 3

   From quantisation: v = nh/2πmr, substitute in force equation

4. 4

   Solve for r: r = n²h²ε₀/πme²Z = n²(0.529Å)/Z

Steps

1. 1

   Consider a conductor of cross-sectional area A with n free electrons per unit volume

2. 2

   Under applied electric field E, electrons drift with average velocity v_d

3. 3

   In time Δt, all electrons in volume A·v_d·Δt cross any cross-section

4. 4

   Number of such electrons: ΔN = n × A × v_d × Δt

5. 5

   Charge transported: ΔQ = eΔN = neAv_dΔt

6. 6

   Current I = ΔQ/Δt = neAv_d

Steps

1. 1

   Two long parallel wires carry currents I₁ and I₂, separated by distance d

2. 2

   Wire 1 sets up magnetic field at the location of wire 2: B₁ = μ₀I₁/2πd (by Ampere's law)

3. 3

   Force on length L of wire 2 due to B₁: F = I₂LB₁ = μ₀I₁I₂L / 2πd

4. 4

   Force per unit length: F/L = μ₀I₁I₂ / 2πd

5. 5

   By right-hand rule: same-direction currents → attractive; opposite-direction → repulsive

Steps

1. 1

   Rectangular coil: N turns, length l, breadth b, current I, in uniform field B

2. 2

   Force on each arm of length l (perpendicular to B): F = BIl

3. 3

   The two forces on opposite arms of length l are equal, opposite, and non-collinear — they form a couple

4. 4

   Perpendicular distance between the couple = b sinθ (θ = angle between plane of coil and B)

5. 5

   Torque = F × b sinθ = BIl × b sinθ = BIA sinθ (A = lb)

6. 6

   For N turns with magnetic moment M = NIA: τ = BINA sinθ = MB sinθ

Steps

1. 1

   Solenoid: length l, cross-section area A, n turns per unit length, total turns N = nl, current I

2. 2

   Magnetic field inside: B = μ₀nI

3. 3

   Flux through each turn: Φ₁ = BA = μ₀nIA

4. 4

   Total flux linkage: NΦ = (nl)(μ₀nIA) = μ₀n²AlI

5. 5

   By definition of self-inductance NΦ = LI → L = μ₀n²Al

Steps

1. 1

   In a series LCR circuit, the same current I flows through R, L and C

2. 2

   Phasor diagram: V_R is along current direction; V_L leads current by 90°; V_C lags current by 90°

3. 3

   V_L and V_C are 180° out of phase — net reactive phasor = V_L − V_C

4. 4

   Resultant voltage phasor: V = √(V_R² + (V_L − V_C)²)

5. 5

   Substituting V_R = IR, V_L = IX_L, V_C = IX_C: V = I√(R² + (X_L − X_C)²)

6. 6

   Impedance Z = V/I = √[R² + (X_L − X_C)²]

7. 7

   At resonance: X_L = X_C → Z_min = R and current is maximum

Steps

1. 1

   Rectangular coil: N turns, area A, rotating at angular velocity ω in uniform magnetic field B

2. 2

   At time t the angle between the normal to the coil and B is θ = ωt

3. 3

   Instantaneous magnetic flux: Φ = NBA cosωt

4. 4

   By Faraday's law: ε = −dΦ/dt = −NBA × d(cosωt)/dt = NBAω sinωt

5. 5

   Peak EMF: ε₀ = NBAω; so ε = ε₀ sinωt

6. 6

   This sinusoidal output is the basis of the AC generator (alternator)

Steps

1. 1

   Two slits S₁ and S₂ separated by distance d; screen at distance D (D >> d)

2. 2

   For point P at height y from centre O: path difference = S₂P − S₁P

3. 3

   Geometric approximation (D >> y, d): path difference Δ ≈ yd/D

4. 4

   Bright fringe condition: Δ = nλ → yₙ = nλD/d

5. 5

   Dark fringe condition: Δ = (2n−1)λ/2 → yₙ = (2n−1)λD/2d

6. 6

   Fringe width β = y_{n+1} − yₙ = λD/d (same for bright and dark fringes)

Steps

1. 1

   Let tan⁻¹x = α and tan⁻¹y = β, so tanα = x and tanβ = y

2. 2

   tan(α+β) = (tanα + tanβ)/(1 − tanα tanβ) = (x+y)/(1−xy)

3. 3

   Therefore α + β = tan⁻¹[(x+y)/(1−xy)]

4. 4

   i.e., tan⁻¹x + tan⁻¹y = tan⁻¹[(x+y)/(1−xy)] when xy < 1

Steps

1. 1

   Start from product rule: d/dx(uv) = u(dv/dx) + v(du/dx)

2. 2

   Integrate both sides: uv = ∫u(dv/dx)dx + ∫v(du/dx)dx

3. 3

   Rearrange: ∫u(dv/dx)dx = uv − ∫v(du/dx)dx

4. 4

   In shorthand: ∫u dv = uv − ∫v du

Steps

1. 1

   Statement: state the three conditions precisely

2. 2

   Geometric meaning: if a smooth curve has equal values at endpoints, the tangent is horizontal at some intermediate point

3. 3

   Proof idea: By extreme value theorem, f has max M and min m on [a,b]

4. 4

   If M = m, f is constant ⟹ f'(c) = 0 for all c

5. 5

   If M ≠ m, at least one extreme occurs at interior point c ⟹ f'(c) = 0

Steps

1. 1

   Conditions: f continuous on [a,b], differentiable on (a,b)

2. 2

   Define auxiliary function g(x) = f(x) − f(a) − [(f(b)−f(a))/(b−a)] × (x−a)

3. 3

   Verify: g(a) = 0 and g(b) = 0; g is continuous on [a,b] and differentiable on (a,b)

4. 4

   Apply Rolle's Theorem to g: ∃ c ∈ (a,b) such that g'(c) = 0

5. 5

   g'(x) = f'(x) − [f(b)−f(a)]/(b−a) = 0 at x = c

6. 6

   Therefore f'(c) = [f(b)−f(a)]/(b−a)

Steps

1. 1

   Standard form: dy/dx + Py = Q, where P and Q are functions of x only

2. 2

   Choose Integrating Factor (IF) = e^(∫P dx)

3. 3

   Multiply both sides by IF: IF·(dy/dx) + P·IF·y = Q·IF

4. 4

   Recognise LHS as d/dx[y × IF] (by product rule, since d(IF)/dx = P·IF)

5. 5

   Equation becomes: d/dx[y × IF] = Q × IF

6. 6

   Integrate both sides: y × IF = ∫(Q × IF) dx + C

Steps

1. 1

   Let A₁, A₂, …, Aₙ be mutually exclusive and exhaustive events (partition of sample space S)

2. 2

   By conditional probability: P(Aᵢ|B) = P(Aᵢ ∩ B) / P(B)

3. 3

   Numerator: P(Aᵢ ∩ B) = P(Aᵢ) × P(B|Aᵢ) — multiplication rule

4. 4

   Denominator: P(B) = Σⱼ P(Aⱼ ∩ B) = Σⱼ P(Aⱼ)·P(B|Aⱼ) — total probability theorem

5. 5

   Substitute both: P(Aᵢ|B) = P(Aᵢ)·P(B|Aᵢ) / Σⱼ P(Aⱼ)·P(B|Aⱼ)

Steps

1. 1

   Let I = ∫₀ᵃ f(x) dx

2. 2

   Substitute x = a − t, so dx = −dt

3. 3

   New limits: when x = 0, t = a; when x = a, t = 0

4. 4

   I = ∫ₐ⁰ f(a − t)(−dt) = ∫₀ᵃ f(a − t) dt

5. 5

   Since t is a dummy variable, replace t with x: I = ∫₀ᵃ f(a − x) dx

6. 6

   Therefore ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a − x) dx